U.U.D.M. Project Report 2014:17

Examensarbete i matematik, 15 hpHandledare och examinator: Karl-Heinz FieselerMaj 2014

Department of MathematicsUppsala University

Construction of Irreducible Polynomials over Finite Fields

Gustav Hammarhjelm

Construction of irreducible polynomials over finitefields

Gustav Hammarhjelm

May 22, 2014

Contents1 Introduction 3

2 Basic results on finite fields 42.1 The reciprocal of a polynomial . . . . . . . . . . . . . . . . . . . . . . . 72.2 The Mobius inversion formula . . . . . . . . . . . . . . . . . . . . . . . 8

3 Finding irreducible polynomials (examples) 9

4 Sequences of irreducible polynomials 124.1 The Q-transformation and the trace . . . . . . . . . . . . . . . . . . . . . 124.2 Sequences of irreducible polynomials over finite fields of characteristic 2 154.3 Sequences of irreducible polynomials over finite fields of odd characeristic 184.4 The polynomial xqn+1 − 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 23

References 25

1

Abstract

In this paper we investigate some results on the construction of irreducible poly-nomials over finite fields. Basic results on finite fields are introduced and proved.Several theorems proving irreducibility of certain polynomials over finite fields arepresented and proved. Two theorems on the construction of special sequences ofirreducible polynomials over finite fields are investigated in detail.

Acknowledgements

I would like to thank my supervisor Karl-Heinz Fieseler for guidance, inspirationand insightful comments. I would also like to thank my family for their support.

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1 IntroductionThe concept of a prime number is well known. The properties that make prime numbersinteresting include (but are not limited to) the fact that a prime number does not admitany non-trivial factorization in integers and that if a prime number divides a product ofnumbers, it necessarily divides one of the factors. The first quality is what defines anirreducible element in any unital ring:

Definition 1.1. Let R be a commutative ring with unity and let r ∈ R. A non-zero, non-unit r is said to be irreducible if r = ab for a, b ∈ R implies a is a unit or b is a unit.

If one is challenged to find, explicitly, infinite sequences of distinct irreducible ele-ments of a ring one can have various outcomes:

In the ring Z the irreducible elements are ±p where p is any prime number. As oftoday, as far as I know, nobody has come up with an explicit infinite sequence of distinctprime numbers.

The challenge turns out to be a rather modest one in some rings. For instance in Q[x],the polynomial ring over the field of rational numbers, it is very easy to explicitly definesequences of irreducible elements, e.g. the sequence xn − 2 where n is a non-zero naturalnumber, using Eisenstein’s criterion.

In this text we will consider the setting when R is the polynomial ring Fq[x] over afinite field Fq. A non-constant polynomial f (x) of Fq[x] is called irreducible over Fq iff (x) = g(x)h(x) for polynomials g(x), h(x) ∈ Fq[x] implies g(x) or h(x) is a unit, i.e.g(x) or h(x) is in Fq, according to the definition of irreducibility. We will show that itis indeed possible (but requires more work than in Q[x]) to generate infinite sequencesof irreducible elements of strictly increasing degrees over Fq[x] for various finite fieldsFq[x].

The existence of such sequences are not only valuable for recreational purposes, theymay also be used for applications in mathematics. Indeed, one important role of irre-ducible polynomials is that one can explicitly construct fields using irreducible polyno-mials through factor rings. If one wants to make explicit calculations in say a finite field,it is often required to find an irreducible polynomial, in order to get information of thestructure of the field. This is important for applications of field theory, for instance errorcorrecting codes.

In this text we shall, after presenting some auxiliary results, investigate some waysof recognizing irreducible polynomials over finite fields. In the last part, we carefullyinvestigate a theorem on the construction of infinite sequences of irreducible polynomialsof increasing degree over finite fields.

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2 Basic results on finite fieldsFirstly, some notation that will be used in the text. If F and K are fields F > K expressesthat F is a field extension of K (or K is a subfield of F). If the extension is finite, then[F : K] denotes the dimension of F over K, when F is considered a vector space overK. If α1, . . . , αn are algebraic over F then F(α1, . . . , αn) is the extension of F obtained byadjoining α1, . . . , αn to F. F[x] denotes the polynomial ring over F and Fq denotes thefinite field of q elements, F∗q its multiplicative group.

Some fundamental results of algebra shall be used frequently, but will not be provedhere, for instance that there is a finite field of pn elements for each prime p and eachpositive natural number n, unique up to isomorphism, as well as the tower law for finiteextensions and that the multiplicative group of a finite field is cyclic.

Theorem 2.1. Let F be a finite field of characteristic p. Then

(a + b)pn= apn

+ bpn, (a − b)pn

= apn− bpn

for a, b ∈ F, n ∈ N>0.

Proof. By the binomial theorem for commutative rings

(a + b)p =

p∑k=0

(pk

)akbp−k

where each p |(

pk

)for each 0 < k < p so (a + b)p = a + b. Now (a + b)pn

= ((a + b)pn−1)p

and the first result follows by induction. For the second result

(a − b)pn= (a + (−b))pn

= apn+ (−b)pn

.

Now if p is odd, (−1)pn= −1, if p is even −1 = 1 so in either case we have obtained the

other result. �

Theorem 2.2. Let Fq be a finite field and let f ∈ Fq[x] be irreducible over Fq, deg f = n.Then the splitting field of f is Fqn . Furthermore, if α is a zero of f , then the other zeros off are given by αq, . . . , αqn−1

.

Proof. The theorem is trivial if n = 1 so assume n > 1. Let α be a zero in the splittingfield of f , α , 0 since f (x) irreducible. [Fq(α) : Fq] = n so Fq(α) � Fqn . Now, supposef (x) =

∑nk=0 akxk, so that f (α) =

∑nk=0 akα

k = 0. By theorem 2.1, for 0 < i < n

0 =

n∑k=0

akαk

qi

=

n∑k=0

aqi

k αkqi

= f (αqi),

since aqk = ak, as ak ∈ Fq. It remains to show that αqi

= αq j, 0 ≤ i, j < n implies i = j (so

that we really have obtained n distinct zeros of f ), until we, with clear conscience, maydeclare Fq(α) � Fqn the splitting field of f .

To this end, we use the fact that an irreducible polynomial f (x) of degree m over afinite field Fq divides xqn

− x if and only if m | n. If m | n then Fqm < Fqn and as Fqn consistsof the zeros of xqn

− x each zero of f (x) is a zero of xqn− x so f (x) divides this polynomial.

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Conversely, if f (x) | xqn− x and β is a zero of f (x) in Fqm , we have the equality

Fqm = F(β) since f (x) is irreducible of degree m, then α is a zero of xqn− x as well and

thus α ∈ Fqn . Therefore we haveFq < Fqm < Fqn

and m | n by the tower law of finite field extensions.Now, for a contradiction, assume αqi

= αq j, 0 ≤ i, j < n. Then, since α , 0 we have

αqi= αq j

⇐⇒ αqi(q j−i−1) = 1 ⇐⇒ (αq j−i−1)qi= 1

by raising the right hand side to the power qn−i and multiplying with α we get αq j−i= α

since αq j−i−1 ∈ Fqn . Thus, α is a zero of x j−i − x and so m | j − i with 0 < j − i < m whichis absurd. �

Remark 2.3. We have seen in the proof of the last theorem that an irreducible polynomialover a finite field of degree m must have m distinct zeros. With this information we candeduce that polynomials of certain forms are never irreducible.

Let Fq be a field of characteristic p and consider the polynomial xp+a for some a ∈ Fq.Let α be a zero of xp + a = 0 with α ∈ Fqp . Then (x − α)p = xp − αp = xp + a and wesee that the only zero of xp + a = 0 is α and since p > 1 the polynomial xp + a must bereducible over Fq since if it would be irreducible, it would have p distinct zeros.

Definition 2.4. Let F be a field and K be a subfield of F. An automorphism σ of F is anautomorphism of F over K if σ(a) = a for all a ∈ K.

Theorem 2.5. Let Fq and Fqm , m > 1 be finite fields. Then the automorphisms of Fqm overFq are precisely σi, i = 1, . . . ,m where σi(α) = αqi

for all α ∈ Fqm .

Proof. That σi are indeed automorphisms of Fqm over Fq is easily seen.Suppose ϕ is an automorphism Fqm over Fq. Let θ be a generator of the multiplica-

tive group of Fqm . If we can determine the image of θ, we determine the automorphismcompletely. ϕ is a linear mapping of Fqm viewed as a vector space over Fq. Now letf be the minimal polynomial of θ over Fq, deg f = m. Since ϕ is linear, we have0 = ϕ( f (θ)) = f (ϕ(θ)). By theorem 2.2 ϕ(θ) = θqk

for some k ∈ {1, . . . ,m} and sothe result follows. �

Remark 2.6. If K is a field and F a finite extension of K, then the extension is callednormal if [F : K] = |Aut(F/K)|, where Aut(F/K) is the group of automorphisms of Fover K. Such extensions are of great importance in Galois theory. By the above theorem,we see that a finite extension of a finite field is always normal.

Later in the text we shall need the concept of a normal basis of a finite field over asubfield. The definition of this concept is presented here.

Definition 2.7. Let F = Fqm and K = Fq be finite fields. A normal basis of F over K is abasis of F over K of the form {α, αq, . . . , αqm−1

} for some α ∈ F.

Remark 2.8. In terms of our above automorphisms σi(α) = αqiof F over K, 0 ≤ i < m,

a normal basis of F over K is a basis of F over K of the form {α, σ1(α), . . . , σm−1(α)} forsome α ∈ F.

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The two following theorems are used to prove that every finite field has a normal basisover any subfield. First a result from linear algebra, whose proof will be omitted, but canbe found in [4]:

Theorem 2.9. Let F be a field and V a finite dimensional F-vector space, dim V = n,and let T : V → V be a linear map. V is T-cyclic, i.e. there is a basis of V of the form{v,T (v) . . . ,T n−1(v)} for some v ∈ V if and only if the characteristic polynomial χT of Tequals the minimal polynomial µT of T .

The following result is stated and proven in [3].

Theorem 2.10. [3] Let G be a group. Let ϕ1, . . . , ϕm be distinct homomorphisms from Gto F∗q and let a1, . . . , am ∈ Fq, not all zero. Then ϕ1, . . . , ϕm are linearly independent i.e.there exists g ∈ G s.t.

a1ϕ1(g) + . . . + amϕm(g) , 0.

The theorem that shows that every finite field has a normal basis over any subfieldfollows beautifully from the last two results.

Theorem 2.11. [3] Let F = Fqm and K = Fq be finite fields. Then there exists a normalbasis of F over K.

Proof. We consider the automorphisms σi(α) = αqiof F over K, where 0 ≤ i < m. These

are m distinct group homomorphisms from F∗ to F∗. Furthermore, σi are linear maps ofF considered as a vector space over K. The statement that F has a normal basis over Kis equivalent with saying that F is σ1-cyclic. We therefore investigate the minimal andcharacteristic polynomial of σ1, denoted µ and χ respectively.

f (x) = xm − 1 clearly satisfies f (σ1) = 0 ∈ End(F). We now show that there is nopolynomial g(x) of degree less than m such that g(σ1) = 0 ∈ End(F). To this end, letg(x) , 0 be given, deg g < m. Then, g(σ1) assumes the form

a0σ01 + a1σ

11 + . . . + am−1σ

m−11 = a0σ0 + a1σ1 + . . . + am−1σm−1

where a0, . . . , am−1 ∈ F are not all zero, and we may apply the previous theorem to con-clude that there is a ∈ F s.t. g(σ1)(a) , 0 and thus g(σ1) , 0. We may now conclude thatthe minimal polynomial µ of σ1 is of degree m. Since χ is of degree m, both are monic,and µ | f (x), µ | χ we must have µ = χ = f .

By theorem 2.9 we know that F is σ1-cyclic, when viewed as a vector space over K,i.e. there is a ∈ F such that {a, σ1(a), . . . , σm−1

1 (a)} is a basis of F over K. This is thedesired normal basis of F over K. �

The following is a well-known test for determining whether an element α of a field Fq

of odd characteristic is a quadratic residue or not, i.e. whether it exists β ∈ Fq with β2 = αor not, and will be used later in the text:

Theorem 2.12. Let F be a finite field of odd characteristic, |F| = q. Then α ∈ F∗ is aquadratic non-residue of F if and only if α(q−1)/2 = −1.

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Proof. For any non-zero α,(α(q−1)/2

)2− 1 = 0 so α(q−1)/2 = ±1, since in a field x2 − 1 = 0

has only two solutions. Let θ be a generator of the multiplicative group of F, then α = θk

for some natural number k. If k is even, then α = (θk/2)2 is a quadratic residue in F andα(q−1)/2 = θk·(q−1)/2 = (θk/2)q−1 = 1.

If on the other hand α = θ2k+1 for some k, α is a quadratic non residue in F, then

α(q−1)/2 = (θ2kθ)(q−1)/2 = (θk)q−1θ(q−1)/2 = −1,

since θ(q−1)/2 must be −1 as θ is a generator of the multiplicative group of F. �

Remark 2.13. A consequence of this theorem is that a non-zero element of a finite fieldof odd characteristic is a quadratic non-residue if and only if it is an odd power of thegenerator of the multiplicative group. From this it can be derived that the product of aquadratic non-residue and a non-zero quadratic residue is again a quadratic non-residue,as well as that the product of two quadratic residues is again a quadratic residue and finallythat the product of two quadratic non-residues is a quadratic residue.

Furthermore, it is seen that the number of non-zero quadratic residues equals the num-ber of quadratic non-residues which is (q−1)/2, which is seen through the characterizationof non-zero elements as odd or even powers of the generator of the multiplicative group.

2.1 The reciprocal of a polynomialLater in the text, especially when constructing sequences of irreducible polynomials, themajor part of the polynomials dealt with will consist of self-reciprocal polynomials, anotion which will be defined here.

Definition 2.14. The reciprocal of a non-zero polynomial f (x) =∑n

k=0 akxk ∈ Fq[x] ofdegree n, denoted f ∗, is the polynomial f ∗(x) =

∑nk=0 akxn−k. A polynomial is called

self-reciprocal if f ∗(x) = f (x).

For a polynomial f (x) ∈ Fq[x] we will often denote f ∗(x) by xn f (1/x), which is tobe intepreted as an element of Fq(x), the field of quotients of Fq[x]. Upon calculation inFq(x) one indeed finds that f ∗(x) = xn f (1/x) and so xn f (1/x) ∈ Fq[x].Remark 2.15. Here follows some remarks about reciprocal polynomials:

1. If f (x) =∑n

k=0 akxk is self-reciprocal then there is a symmetry in the coefficients off i.e. ak = an−k for k = 0, . . . , n. The converse holds as well.

2. For f (x), g(x) ∈ Fq[x] we have

( f g)∗ = xdeg f g f (1/x)g(1/x) = xdeg f f (1/x)xdeg gg(1/x) = f ∗g∗

and in particular (c f )∗ = c f ∗ for c ∈ Fq. If f (0) , 0 then ( f ∗)∗ = f .

3. Let f ∈ Fq[x] be irreducible over Fq, f (0) , 0. Then f ∗ is irreducible over Fq.

Proof. Since f (0) , 0 then deg( f ) = deg( f ∗) and so ( f ∗)∗ = f . Suppose f ∗ = gh.Then f = ( f ∗)∗ = (gh)∗ = g∗h∗ which implies that g∗ or h∗ is constant. Supposew.l.o.g. that h∗(x) = xdeg(h)h(1/x) is constant. Thus we must have that h(x) = axn

for some a ∈ Fq, n ∈ N. n > 0 would imply that f ∗(0) = 0 which leads to deg( f ) <deg( f ∗), contradiction. Thus n = 0 and f ∗ is irreducible. �

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2.2 The Mobius inversion formulaDefinition 2.16. Let ω(n) be the arithmetic function with ω(n) =

∑p|n 1, so that ω(n) is

the number of distinct primes that divide n. Now set

µ(n) =

{(−1)ω(n) if n is square free,0 otherwise.

Theorem 2.17. [7]. µ(n) is a multiplicative arithmetic function and∑

d|n µ(d) is 0 if n > 1and 1 if n = 1.

Theorem 2.18. The Mobius inversion formula part I, [7]. If F, f are arithmetic functionswith F(n) =

∑d|n f (d) for all n ∈ N>0 then f (n) =

∑d|n µ(d)F(n/d) for all n ∈ N>0.

Proof. Let I be the set of ordered pairs (a, b) with ab|n.∑d|n

µ(d)F(n/d) =∑d|n

µ(d)∑

e|(n/d)

f (e) =∑d|n

∑e|(n/d)

µ(d) f (e)

=∑

(d,e)∈I

µ(d) f (e) =∑e|n

f (e)∑

d|(n/e)

µ(d) = f (n),

since∑

d|n µ(d) is 0 if n > 1 by the previous theorem. �

Theorem 2.19. The Mobius inversion formula part II, [7]. If F, f are arithmetic functionswith f (n) =

∑d|n µ(d)F(n/d) for all n ∈ N>0 then F(n) =

∑d|n f (d) for all n ∈ N>0.

Example 2.20. Let ϕ : N→ C be Euler’s totient function. Then ϕ(n)n =

∑d|n

µ(d)d .

Proof. One can verify that∑

d|n ϕ(d) = n =: F(n) for all n ∈ N (by for instance observingthat ϕ is multiplicative and that the identity holds for prime powers). Set ϕ(n) = f (n).Applying a Mobius inversion to the identity∑

d|n

f (d) =∑d|n

ϕ(d) = n = F(n)

yields the desired result. �

Theorem 2.21. Let Fq be a finite field. Let Iq(n) denote the number of irreducible monicpolynomials over Fq of degree n. Then

Iq(n) =1n

∑d|n

µ(d)qn/d.

Proof. Let n be given. Form the polynomial g(x) = xqn− x, whose splitting field is Fqn

since all elements of Fqn are zeros of g(x). Let f be monic, irreducible over Fq of degreed | n. Fqd is the splitting field of f by theorem 2.2. Since d | n, Fqd is a subfield of Fqn andthus the zeros of f are contained in the set of zeros of g and consequently f | g since allzeros of f are simple.

Now let f | g, f monic and irreducible over Fq. Now it is demonstrated that deg f =

d | n. If α is a zero of f then αqd−1 = 1 since the splitting field of f is Fqd . Since f | g thezeros of f are contained in the set of zeros of g and therefore Fqd is a subfield of Fqn andso d | n.

We have now demonstrated that xqn− x is the product of all irreducible monic poly-

nomials over Fq with degrees dividing n. Therefore we have qn =∑

d|n Iq(d)d. SettingF(n) = qn, f (n) = nIq(n) and applying a Mobius inversion yields the desired result. �

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3 Finding irreducible polynomials (examples)Here are some examples of how one could go about finding the elusive irreducible poly-nomials.

Theorem 3.1. Let p be a prime. The polynomial f (x) = xp − x + a ∈ Fq[x], q = pn withn ≥ 1, is irreducible over Fq if and only if it has no zeros in Fq.

Proof. Let α be a zero of f in some extension field of Fq. Since for all b ∈ Fp, b is a zeroof xp − x, and by theorem 2.1, α + b is a zero of f for every b ∈ Fp. These are all zerosof f . Thus, the splitting field of f is Fq(α). Let p(x) be an irreducible factor of f (overFq) so that Fq[x]/(p(x)) is a field. Then p(α + b) = 0 for some (possibly several) b ∈ Fp

and we must have Fq[x]/(p(x)) � Fq(α). Thus, for any irreducible factor p of f we haveFq[x]/(p(x)) � Fq(α), which implies that all irreducible factors have the same degree. Ifthe number of irreducible factors are k and each is of degree n then we must have kn = p.But, by assumption, f has no zeros in Fq. Therefore n > 1 and we must have k = 1 so fis irreducible.

The converse is trivial. �

Granted theorem 3.1 we can easily establish that the representation of a finite field asa factor ring is not unique. For instance, if p = 5, then both x5 − x + 1, x5 − x + 2 areirreducible over F5 so we have

F5[x]/(x5 − x + 1) � F5[x]/(x5 − x + 2) � F55 .

The following theorem unveils an interesting way of forging two irreducible polyno-mials, yielding another irreducible polynomial of higher degree:

Theorem 3.2. Let F be a finite field and let f , g ∈ Fq[x] be irreducible over Fq, wheredeg f = m, deg g = n, gcd(m, n) = 1 and m, n > 1. Then the polynomials

h×(x) =∏

f (α)=0

∏g(β)=0

(x − αβ), h+(x) =∏

f (α)=0

∏g(β)=0

(x − (α + β))

are irreducible over Fq of degree mn, where the products range over all zeros of f , g inthe splitting fields of f and g.

Proof. The statement is proved for h×(x), the proof for h+(x) works analogously. Let α1

be a zero of f in the splitting field Fqm of f and let β1 be a zero of f in the splittingfield Fqn of g. (α1β1)qm−1 = β

qm−11 ∈ Fq(α1β1) since α1 ∈ Fqm by theorem 2.2. Thus

(βqm−11 )qr−m

= βqr−qr−m

1 = β−qm−r

1 ∈ Fq(α1β1) where r is chosen so that r is a multiple ofn greater than m. We claim that m − r is not a multiple of n since otherwise n | m,contradiction. Thus (β−qr−m

1 )−1 = βqr−m

1 is a zero of g by theorem 2.2 belonging to Fq(α1β1)and therefore β1 ∈ Fq(α1β1) by the same theorem and consequently α1 ∈ Fq(α1β1) . ThusFq < Fq(α1) < Fq(α1β1). By the tower law for finite field extensions we have

[Fq(α1β1) : Fq] = [Fq(α1β1) : Fq(α1)][F(α1) : Fq]

so m = deg(α1,Fq)| deg(α1β1,Fq) and in the same way n = deg(β1,Fq)| deg(α1β1,Fq).Since m, n are relatively prime mn| deg(α1β1,Fq) and since

mn ≥ [Fq(α1, β1) : Fq] ≥ [Fq(α1β1) : Fq] = deg(α1β1,Fq)

9

we have mn = deg(α1β1,Fq). Now, if we can show h×(x) ∈ Fq[x], h×(x) must be irre-ducible since deg h× = mn and it has α1β1 as a zero. First, observe that Fq(α1) ∩ Fq(β1) =

Fq, for if the intersection were greater, with an element γ < Fq then deg(γ,Fq) > 1 and

deg(γ,Fq)| deg(α1,Fq) = m, deg(γ,Fq)| deg(β1,Fq) = n

contradicting the fact that m, n are relatively prime. Now

h×(x) =∏

f (α)=0

∏g(β)=0

(x − αβ) =∏

g(β)=0

∏f (α)=0

β(β−1x − α) =∏

g(β)=0

βm f (β−1x),

since f (x) =∏

f (α)=0(x − α). Thus h×(x) ∈ (Fq(β1))[x], as Fq(β1) is the splitting field ofg, by theorem 2.2. In a similiar manner, one finds that h×(x) ∈ (Fq(α1))[x] so h×(x) ∈(Fq(α1) ∩ Fq(β1))[x] = Fq[x]. �

Example 3.3. Let f (x) = x2 + x+1, g(x) = x3 + x+1 ∈ F2[x]. These polynomials fulfil thehypothesis of the last theorem. We find their composition h×(x). We see from the proofof the theorem that

h×(x) =∏

f (α)=0

α3g(α−1x)

where the product ranges over the zeros of f (x) in F4, call them α1, α2. With this notationwe find

h×(x) = (x3 + α21x + α3

1)(x3 + α22x + α3

2),

and after further simplification

h×(x) = x6 + (α21 + α2

2)x4 + (α31 + α3

2)x3 + (α21α

22)x2 + (α2

1α32 + α3

1α22)x + α3

1α32.

Since f (x) = (x + α1)(x + α2) = x2 + x + 1 we obtain α1α2 = 1, α1 + α2 = 1. Using theseidentities, we eventually find

h×(x) = x6 + x4 + x2 + x + 1,

irreducible over F2 of degree 6.

We conclude this section by presenting a way of how one can obtain new irreduciblepolynomials from given ones via automorphisms.

Let Fq be a finite field and let σ be an automorphism of Fq. Given such σ define

σ : Fq[x]→ Fq[x]

as

σ( f (x)) =

n∑k=0

σ(ak)xk

where f (x) =∑n

k=0 akxk ∈ Fq[x]. We claim that σ is a homomorphism. Obviously, forpolynomials f (x), g(x) we have σ( f (x) + g(x)) = σ( f (x)) + σ(g(x)). We now verify thatσ( f (x)g(x)) = σ( f (x))σ(g(x)) when g(x) is a monomial axn and the claim follows sinceσ is an additive homomorphism. Suppose f (x) =

∑nk=0 akxk, then

σ( f (x)axn) = σ

n∑k=0

a · akxk+n

=

n∑k=0

σ(a·ak)xk+n = σ(a)xnn∑

k=0

σ(ak)xk = σ( f (x))σ(axn).

10

This shows that σ indeed is a homomorphism. Furthermore, σ is an isomorphism, sinceit has an inverse given by

σ−1= σ−1.

We can now state and prove a theorem on how one might produce new irreducible poly-nomials from known ones through automorphisms.

Theorem 3.4. A polynomial f (x) ∈ Fq[x] is irreducible over Fq if and only if σ( f (x)) isirreducible over Fq, where σ is defined as above using any automorphism σ : Fq → Fq.

Proof. Suppose f (x) is reducible, f (x) = g(x)h(x), deg g, deg h > 0, then, since

σ( f (x)) = σ(g(x)h(x)) = σ(g(x))σ(h(x))

and σ is an automorphism, in particular σ(a) , 0 for a , 0, deg g = degσ(g) anddeg h = degσ(h) which shows that σ( f (x)) is reducible.

If σ( f (x)) is reducible for some σ then as above f (x) = σ−1(σ( f (x))) is reducible. �

Now, if we start with fields F = Fqm > Fq = K and a non-trivial automorphismof F over K, for instance, σi(α) = αqi

for 1 ≤ i < m and an irreducible polynomialf (x) ∈ F[x] having not all coefficients in the set F′ ⊂ F of elements left fixed by σi, thenwe end up with a new irreducible polynomial given by σi( f (x)). Later in the text we shallillustrate this by giving an example of how the latest theorem can be used to generate anew sequence of irreducible polynomials from a given one.

11

4 Sequences of irreducible polynomialsIn this section a theorem on the construction of certain sequences of irreducible polyno-mials over finite fields shall be studied in detail. The goal is to present a proof of thefollowing theorem:

Theorem 4.1. Let q be the power of an odd prime p and let f1 ∈ Fq be monic andirreducible over Fq[x] of degree m, with m even if p ≡ 3 mod 4, such that f1(1) f1(−1) isnot a quadratic residue of Fq, then the monic polynomials defined recursively by

fn+1(x) = (2x)m2n−1fn

(x2 + 1

2x

)are all irreducible over Fq.

This theorem is a slight modification of a theorem presented on page 45 of [8]. Thestatement of the theorem in [8] is identical to the one above, with the exception that theassumption m even if p ≡ 3 mod 4 is dropped. As we will later see in this text, thisassumption cannot be omitted.

Theorem 4.1 was proven in [2] by S.D. Cohen who expanded on results obtained byH. Meyn in [5]. We shall follow the approaches of these documents closely in this section,and study the arguments used in detail, in order to achieve a proof of theorem 4.1.

Firstly, we shall introduce important concepts used in both papers, before presentinga way of constructing sequences of irreducible polynomials over fields of characteristic2 of growing degree. This will be followed by a section devoted to the proof of theorem4.1, which concerns finite fields of odd characteristic.

4.1 The Q-transformation and the traceOne fruitful approach in the quest of finding sequences of irreducible polynomials ispresented in [5]. The idea is to look at a certain transformation Q : Fq[x]→ Fq[x], Q( f ) =

f Q and determine conditions under which there is an inheritance of irreducibility when fis transformed to f Q. The transformation is as follows:

Definition 4.2. Let f ∈ Fq[x] be a polynomial. Let f Q(x) = xdeg f f (x + 1/x) interpreted asan element in Fq(x), but actually an element of Fq[x]. More precisely, if f (x) =

∑nk=0 akxk,

an , 0, then f Q(x) =∑n

k=0 ak(x2 + 1)kxn−k. The mapping f (x) 7→ f Q(x) will occasionallybe referred to as the Q-transform.

Remark 4.3. Note that if deg f = n then deg f Q = 2n and that ( f Q)∗ = f Q, i.e. f Q isself-reciprocal.

Furthermore, if f ∈ Fq[x] and a ∈ Fq then we see from the definition of f Q(x) that wehave

(a f (x))Q = a f Q(x).

More generally, if f (x) = g(x)h(x), g, h ∈ Fq[x], then

f Q(x) = xdeg f f (x + 1/x) = xdeg gxdeg hg(x + 1/x)h(x + 1/x) = gQ(x)hQ(x),

so that f Q is irreducible only when f is.

12

There is also a correspondence between all polynomials of degree n and all self-reciprocal polynomials of degree n. If we count the polynomials of degree n over Fq

we find that there are exactly qn(q − 1). By the remark following definition 2.14 a poly-nomial

∑2nk=0 akxk is self reciprocal if and only if a2n−k = ak for all 0 ≤ k ≤ 2n. Therefore,

when constructing an irreducible polynomial of degree 2n, the polynomial is determinedby choosing a0, . . . , an with the only restriction a0 , 0. Thus, there are exactly qn(q − 1)self-reciprocal polynomials of degree 2n over Fq. Furthermore, if f is of degree n, asnoted above f Q is self-reciprocal of degree 2n, and, as will now be shown, the mappingf 7→ f Q is injective.

Suppose f Q(x) = gQ(x), f , g ∈ Fq[x]. Clearly f , g must have the same degrees, n say,so in other words, we have

xn f (x + 1/x) = xng(x + 1/x).

Let 0 , β ∈ Fq, the algebraic closure of Fq. In order to show injectivity, it suffices toshow that f (β) = g(β) and that g(0) = f (0). Let α ∈ Fq be a zero of x + 1/x = β ⇐⇒x2 − βx + 1 = 0. Then α , 0 and

f Q(α) = gQ(α) ⇐⇒ αn f (α + 1/α) = αng(α + 1/α) ⇐⇒ f (β) = g(β).

It remains to show that f (0) = g(0), i.e. that the constant terms of f , g agree. But theconstant term of f (x) is the coefficient of the highest term in f Q(x), and likewise for g,and since f Q = gQ, f (0) = g(0), so f (x) = g(x) and f 7→ f Q is injective (actually bijective,since domain and image are finite).

The next theorem plays an important role both in a construction of sequences of irre-ducible polynomials over fields of characteristic 2, as well as in the proof of theorem 4.1,which is our goal to prove. It gives a necessary and sufficient condition for when f Q isirreducible if f is.

Theorem 4.4. (Lemma 5 of [5]). Let f (x) ∈ Fq[x] be irreducible over Fq with deg f = n.Then f Q is irreducible over Fq if and only if g(x) = x2−βx + 1 ∈ Fqn[x] is irreducible overFqn , where β is any zero of f .

Remark 4.5. If x2 − βx + 1 ∈ Fqn[x] is irreducible for some zero β of f (x) then it isirreducible for any other zero of f (x). By theorem 2.2 the other zeros are βq, . . . , βqn−1

which can be expressed in terms of the automorphism σ : Fqn → Fqn of Fqn over Fq givenby σ(α) = αq for α ∈ Fqn as σ(β), . . . , σn−1(β). Therefore x2−σk(β)x+1 are all irreducibleover Fqn for 1 ≤ k ≤ n − 1 by theorem 3.4 which proves the claim.

Proof. Suppose g(x) is irreducible over Fqn . Firstly, we show that 0 is not a zero of f Q.If it would be, then the constant term of f Q would be 0. But the constant term of f Q

is that of xn in f , obviously non-zero. Now, let α , 0 be a zero of f Q. Our aim is toshow that deg(α,Fq) = 2n = deg f Q. Since 0 = f Q(α) = αn f (α + 1/α) we find thatf (α + 1/α) = 0, since α , 0. Let β = α + 1/α, deg(β,Fq) = n. Furthermore g(α) = 0.Since g(x) is assumed to be irreducible over Fqn , Fqn[x]/(g(x)) is a field, isomorphic to(Fq(β))(α) = Fq(α) and we have, by the tower law for finite field extensions

[Fq(α) : Fq] = [Fq(α) : Fq(β)][Fq(β) : Fq] = 2n

13

so deg(α,Fq) = 2n = deg f Q and we have deduced that f Q must be irreducible.If, on the other hand, f Q(x) is irreducible over Fq, and α is a zero of f Q, so that

[Fq(α) : Fq] = 2n, then, by setting β = α+1/α, f (β) = 0, we see that for g(x) = x2−βx+1,g(α) = 0. If g would be reducible α would be a zero of some linear polynomial of Fqn[x]and so α ∈ Fqn contradicting that [Fq(α) : Fq] = 2n. �

Quite a lot of work is dedicated to transforming the above necessary and sufficientcondition for when f Q inherits irreducibility of f to a more applicable one. This is donethrough analyzing the irreducibility of x2 − βx + 1 ∈ Fqn[x] and the analysis dependsstrongly on whether the characteristic of the field is odd or not. So while the abovetheorem holds for any characteristic it will be transformed to give other conditions forinheritance of irreducibility depending on the characteristic of Fq as we shall later see.

In Meyn’s paper [5] the notion of the trace of an element plays an important role in thesearch of sequences of irreducible polynomials over fields of characteristic 2, as it allowsbetter usage of theorem 4.4. This notion is introduced here as it is valid for finite fields ofany characteristic.

Definition 4.6. Let Fqm = F,Fq = K be finite fields. Let α ∈ F. The trace of α over K isdenoted and defined as

TrF/K(α) =

m−1∑k=0

αqk.

Remark 4.7. If α ∈ F > K then TrF/K(α) ∈ K. For let f be the minimal polynomial of αover K. Then deg f = d | m, F = Fqm . By theorem 2.2 the elements α, . . . , αd−1 are thezeros of f . Now, by setting

∏m−1k=1 (x − αqk

) = g(x) = f (x)m/d ∈ K[x], one sees that thesecond highest coefficient of g is −TrF/K(α) so this element must be in K.

Alternatively, we observe that the trace of α ∈ F over K an element of K left invariantby all automorphisms of F over K. Thus TrF/K(α) ∈ K, by the theory of Galois.

In particular, if Fqm = F,Fq = K are finite fields, and the degree of the minimalpolynomial of α over K is equal to m (so that F = K(α)), then −TrF/K(α) equals thecoefficient of xm−1 in f (x).

Some properties of the trace ([3] page 55):

• TrF/K : F → K is linear (F considered a vector space over K).

• TrF/K(αq) = TrF/K(α), for all α ∈ F.

• The trace function is transitive, i.e. if K < F < L are finite fields fields and α ∈ Lthen TrL/K(α) = TrF/K(TrL/F(α)).

14

4.2 Sequences of irreducible polynomials over finite fields of charac-teristic 2

To be able to use theorem 4.4, one utilizes another irreducibility condition for x2−βx+1 ∈F2n[x], which is presented shortly. First, an example from [6] on how the trace can be usedto show irreducibility of quadratic polynomials over fields of characteristic 2.

Theorem 4.8. Let F = F2k , K = F2, and let f (x) = x2 + x + β ∈ F[x]. f (x) has a zero inF, is not irreducible over F, if and only if TrF/K(β) = 0. In other words f (x) is irreducibleover F if and only if TrF/K(β) = 1.

Proof. From theorem 2.11 we get that F has a normal basis over K, i.e. a basis of theform {α2i

: 0 ≤ i < k} for some α ∈ F. So if there is a solution y of f (x) = 0, with y ∈ F,we may write y = αy0 + . . . + α2k−1

yk−1, β = αb0 + . . . + α2k−1bk−1. Now

y2 = (αy0 + . . . + α2k−1yk−1)2 = α2y2

0 + . . . + (α2k−1)2y2

k−1

which is equal to αyk−1 + α2y1 . . . + α2k−1yk−2 since yi ∈ F2 andless (α2k−1

)2 = α2k= α.

By the condition that y2 + y = β, and comparison of coefficients, we obtain y0 + yk−1 =

b0, y1 + y0 = b1, . . . , yk−1 + yk−2 = bk−1. Adding all those equations, we obtain 0 =∑k−1i=0 2yi =

∑k−1i=0 bi. The claim is now that TrF/K(β) =

∑k−1i=0 bi. By linearity of the trace

TrF/K(β) =∑k−1

i=0 biTrF/K(αqi) and by the other property of the trace mentioned in remark

4.7 we have TrF/K(αqi) = TrF/K(α) for all i in the sum. So, it only remains to show that

TrF/K(α) =∑k−1

i=0 αqi

= 1, but this follows since TrF/K(α) ∈ K and so TrF/K(α) = 0 orTrF/K(α) = 1 but the first situation cannot arise since {α, . . . , α2k−1

} is a basis of F over K.Now, suppose TrF/K(β) = 0. Then we can construct solutions y of the equation by

letting y0 = a, y1 = a + b1, y2 = a + b1 + b2, . . . , ym−1 = a + b1 + . . . + bm−1, a = 0, 1, asshown in [6]. �

We now turn to the promised irreducibility condition for g(x) = x2−βx+1 = x2+βx+1over F2n , aided by the last result.

Theorem 4.9. Let K = F2 and 0 , β ∈ F2k = F. Then the equation x2 + βx + 1 = 0 has asolution in F if and only if TrF/K( 1

β) = 0; consequently, x2 + βx + 1 is irreducible over F

if and only if TrF/K( 1β) = 1).

Proof. Suppose the equation x2 + x + 1β

has solutions ξ, η ∈ F, a situation which occursif and only if TrF/K( 1

β) = 0 by the previous theorem. Then, obviously, ξ, η are non-zero,

ξη = 1β, ξ + η = 1, and it is verified that ξ

ηis a solution of x2 + βx + 1 = 0:(

ξ

η

)2

+ β ·ξ

η+ 1 =

ξ2 + βξη + η2

η2 =(ξ + η)2 + 1

η2 = 0.

Thus, if TrF/K( 1β) = 0, then x2 + βx + 1 = 0 has a solution in F.

Now suppose x2 + βx + 1 = 0 has a solution in F. Since β , 0, this equation isequivalent to 1

βx2 + x + 1

β= 0. Suppose this equation has a solution a. Then a

βis a solution

of x2 + x + 1β2 = 0, which implies that TrF/K( 1

β2 ) = 0, by the previous theorem. By thevirtue of a property possessed by the trace 0 = TrF/K( 1

β2 ) = TrF/K( 1β). �

15

The following theorem connects the most recent theorem with theorem 4.4 in order toobtain conditions for when irreducibility of f Q is inherited by the irreducibility of f :

Theorem 4.10. (Theorem 6 of [5]). Let F = F2k , k > 0, K = F2, and let f (x) =

xn +∑n−1

k=0 akxk be irreducible over F. Then f Q(x) is irreducible over F if and only ifTrF/K(a1/a0) = 1.

Proof. Let L = F2nk . Let β be a zero of f , β ∈ L. By the previous theorem and theorem4.4 f Q(x) is irreducible over F if and only if TrL/K( 1

β) = 1. f ∗ is irreducible over F

by remark 2.15 and f ∗(1/β) = 0 (recall that f ∗(x) = xn f (1/x)). Furthermore f ∗(x)/a0 ismonic and irreducible over F of degree n with 1/β as a zero and therefore it is the minimalpolynomial for 1/β over F. By the remark 4.7 we have TrL/F(1/β) is the coefficient of xn−1

in f ∗(x)/a0, namely a1/a0. Since the trace function is transitive, i.e. if K < F < L arefinite fields fields and α ∈ L then TrL/K(α) = TrF/K(TrL/F(α)), we find that

1 = TrL/K(1/β) = TrF/K(TrL/F(1/β)) = TrF/K(a1/a0).

�

Remark 4.11. In F2, given an irreducible polynomial f , f Q is irreducible if and only ifthe linear term of f has coefficient 1. And clearly, in any field of characteristic 2, thelinear term of f must have non-zero coefficient a1 in order for f Q to be irreducible, sinceotherwise TrF/K(a1/a0) = 0, regardless of the value of a0 , 0.

Example 4.12. Let α be a root of x3 + x + 1 ∈ F2[x], so that F8 = F2(α). Then α, as can beverified, is a generator of F∗8. Then consider x + α ∈ F8[x], irreducible. TrF8/F2(1/α) = 1,so f Q(x) is irreducible over F8. However, f Q(x) = x2 +αx + 1 and TrF8/F2(α/1) = 0 so f Q2

must be reducible over F8. Indeed

f Q2= x4 + αx3 + x2 + αx + 1 = (x2 + α5x + α6)(x2 + α6x + α).

In the light of the above example, we request conditions which assure that if a poly-nomial f satisfies the requirements of theorem 4.10, then f Q also satisfies those require-ments. It turns out, that it is sufficient to require an extra property of f , namely self-reciprocality.

Theorem 4.13. Let F = F2k , k > 0, K = F2. If a polynomial f (x) = xn + a1xn−1 + . . . +a1x + 1 ∈ F[x] is self-reciprocal and irreducible over F and satisfies TrF/K(a1) = 1, thenf Q(x) = x2n + b1x2n−1 + . . . + b1x + 1 ∈ F[x] satisfies TrF/K(b1) = 1.

Proof. f being self-reciprocal implies that there is a certain symmetry in its coefficients,namely ak = an−k for k = 0, . . . , n. Now we need only to observe that since f exhibitsthis symmetry in its coefficients the linear terms and constant terms of f Q will be thesame as for f , so b1 = a1. This is readily seen from the definition of f Q, f Q(x) =∑n

k=0 ak(x2 + 1)kxn−k = an + an−1x + . . . = 1 + a1x + . . .. �

Now we have uncovered a weakness of the polynomial in the example preceding thelast theorem, namely that the polynomial under scrutiny failed to be self-reciprocal.

Theorem 4.13 now ascertains that, if we start with a polynomial fulfilling all criteria,we can generate an infinite sequence of irreducible polynomials by repeatedly applyingthe Q-transform, since the assumptions in theorem 4.13 guarantee that theorem 4.10 canbe applied repeatedly.

16

Example 4.14. Here follows some examples of when the theorem can be applied.

1. The simplest self-reciprocal, irreducible polynomial that comes to mind is x + 1 ∈F2k[x]. We calculate the required trace in order to see whether or not theorems 4.10and 4.13 apply, i.e. we must calculate the trace of 1:

TrF2k /F2(1) =

k−1∑i=0

12 j= k

which is 1 if k odd and 0 if k even. So our theorems tell us that the Q-transformapplied to x + 1 yields infinitely many irreducible polynomials over F2k preciselywhen k is odd. The sequence originating from x+1 will be a sequence of irreduciblepolynomials of degree a power of 2 over any field F2k with k odd, and the first fewelements are

x + 1, x2 + x + 1, x4 + x3 + x2 + x + 1, x8 + x7 + x6 + x4 + x2 + x + 1x16 + x15 + x14 + x13 + x12 + x11 + x8 + x5 + x4 + x3 + x2 + x + 1

2. If we consider F4 = F2(α) where α is a zero of x2 + x + 1, irreducible over F2, then

TrF4/F2(α−1) = TrF4/F2(α) = α + α2 = α + α + 1 = 1

and so the theorems apply to the two self reciprocal irreducible polynomials

f1(x) := x2 + αx + 1, f2(x) := x2 + α−1x + 1.

Irreducibility of f1(x), f2(x) holds due to theorem 4.9 and thus we can iterate theQ-transform to those polynomials, and we obtain

f Q1 (x) = x4 + αx3 + x2 + αx + 1,

f Q2

1 (x) = x8 + αx7 + αx6 + αx4 + αx2 + αx + 1.

f Q2 (x) = x4 + α−1x3 + x2 + α−1x + 1,

f Q2

2 (x) = x8 + α−1x7 + α−1x6 + α−1x4 + α−1x2 + α−1x + 1.

3. The only polynomial over F8 of degree 2 on which theorem 4.13 is applicable isx2 + x + 1, since every other self reciprocal polynomial x2 + βx + 1 fails to satisfy

TrF8/F2(β−1) = TrF8/F2(β) = 1

which has to be satisfied in order to apply theorems 4.9 and 4.13.

A natural question now arises, is there always a choice for a polynomial satisfyingtheorem 4.13? This question is partially dealt with in the case of underlying field F2 in thepaper by Meyn, [5], where it is shown that there exists a monic irreducible self reciprocalpolynomial of degree 4m and the linear coefficient is 1 for every odd m.

17

4.3 Sequences of irreducible polynomials over finite fields of odd charac-eristic

While in the case of characteristic 2, the transform f 7→ f Q succeeded in rendering se-quences of irreducible polynomials, when certain conditions were imposed on the initialpolynomial, it turns out that in the case of odd characteristic, we need to modify ourtransform slightly in order to generate such sequences over finite fields of general oddcharacteristic.

If Fq is of odd characteristic, one can obtain an irreducibility criterion for g(x) =

x2 − βx + 1 ∈ Fq[x], in order to use theorem 4.4 as follows. Since g(x) is of degree2 it is irreducible over Fq if and only if it has no zeros in Fq. By rewriting g(x) = 0as (2x − β)2 = β2 − 4, by rearrangements and completion of squares (enabled by oddcharacteristic), we see that g(x) is irreducible over Fq if and only if β2 − 4 is a quadraticnon-residue of Fq.

The above theorem and the preceding irreducibility criterion of x2−βx+1 can be usedto prove the following theorem of [5] (here proved in greater detail):

Theorem 4.15. Let Fq be a finite field of odd characteristic and let f (x) ∈ Fq[x] be anirreducible monic polynomial over Fq of degree n. Then f Q is irreducible over Fq if andonly if f (2) f (−2) is a quadratic non-residue of Fq.

Proof. By theorem 4.4 f Q is irreducible if and only if g(x) = x2 − βx + 1 is irreducibleover Fqn , where β is a zero of f in Fqn (the splitting field of f ). This happens if and only ifβ2−4 is a quadratic non-residue of Fqn , by theorem 2.12 if and only if (β2−4)(qn−1)/2 = −1.

By theorems 2.1 and 2.2, for a ∈ Fq

f (a) =∏

{γ: f (γ)=0}

(a− γ) =

n−1∏k=0

(a− βqk) =

n−1∏k=0

(a− β)qk= (a− β)1+q+...+qn−1

= (a− β)(qn−1)/(q−1),

where the first product ranges over all zeros γ of f in the splitting field of f . Because ofthis we have β2 − 4 is a quadratic non-residue of Fqn ⇐⇒ (β2 − 4)(qn−1)/2 = −1 ⇐⇒((2 − β)(−2 − β))(qn−1)/2 = −1 ⇐⇒

(((2 − β)(−2 − β))(qn−1)/(q−1))(q−1)/2

= −1 ⇐⇒

( f (2) f (−2))(q−1)/2 = −1 ⇐⇒ f (2) f (−2) is a quadratic non-residue of Fq. �

This theorem corresponds to theorem 4.10 in the sense that it provides transformationof the abstract necessary and sufficient condition of when f Q inherits irreducibility fromf given in theorem 4.4 into a more practical one. The condition in 4.10 was to verify atrace property of a certain element in the field of coefficients Fq of the polynomial underconsideration. We succeeded in finding a similar condition here as well, namely to verifythat a certain element in Fq is a quadratic non-residue. This should be compared to thetask of using 4.4 for practical purposes, where one has to determine the irreducibility of aquadratic polynomial over an extension field of Fq.

Using theorem 4.15 we can define our first sequence of irreducible polynomials overfields of odd characteristic.

Example 4.16. Let f (x) = x2 + 2x + 2 ∈ F3[x], which is irreducible. Furthermoref (2) f (−2) = f (−1) f (1) = −1 which is a quadratic non-residue of F3. Now, if f Qn

de-notes repeated application of f 7→ f Q n times, then

f Qn+1(2) f Qn+1

(−2) = 2deg f Qn

(−2)deg f Qn

f Qn(2 + 2−1) f Qn

(−2 + (−2)−1) = f Qn(−2) f Qn

(2),

18

since deg f Qnis even (we define f Q0

:= f ). Thus, by induction and theorem 4.15, thesequence defined by

fn+1(x) = f Qn (x),

with f1(x) = f (x), is a sequence of irreducible polynomials over F3, deg fn = 2n, the firstfew being

f2(x) = x4 + 2x3 + x2 + 2x + 1, f3(x) = x8 + 2x7 + 2x6 + 2x5 + 2x3 + 2x2 + 2x + 1,f4(x) = x16 + 2x15 + x14 + x13 + x12 + x11 + 2x9 + x8 + 2x7 + x6 + x4 + x3 + x2 + 2x + 1.

We note that this process will succeed in general: Let f (x) ∈ F3k[x] be irreducible ofeven degree s.t. f (2) f (−2) = f (−1) f (1) is a quadratic non-residue of F3k . Then f Qn

(x)are irreducible for n ∈ N. This is true because

f Qn+1(2) f Qn+1

(−2) = 2deg f Qn

(−2)deg f Qn

f Qn(2 + 2−1) f Qn

(−2 + (−2)−1) = f Qn(−2) f Qn

(2)

so the claim follows by induction and theorem 4.15.Observe that the induction step heavily depended on that the characteristic was 3.

Here is an example of a situation when repeated application of the Q-transform failsto produce more than one irreducible polynomial regardless of the starting polynomial.

Example 4.17. Let f (x) ∈ F5k[x], where k ∈ N>0, be irreducible, s.t. f (2) f (−2) is aquadratic non-residue of F5k . Thus, f Q is irreducible by theorem 4.15. However,

f Q(2) f Q(−2) = 2deg f (−2)deg f f (2 + 2−1) f (−2 + (−2)−1) = f (0)2

which is a quadratic residue, hence f Q2is not irreducible by theorem 4.15.

In order to prove theorem 4.1 we introduce a new polynomial transformation, involv-ing the Q-transform.

Definition 4.18. Let Fq be a finite field of odd characteristic. Given a polynomial f ∈Fq[x] of degree n, let f R(x) = 2n f Q(2−1x) = (2x)n f (2−1(x + 1/x)). Furthermore, for apolynomial f , let λ( f ) = f (1) f (−1).

Remark 4.19. Looking at f R(x) = 2n f Q(2−1x) we see that the factor 2n exists only for nor-malization purposes. The crucial difference between the R-transform and the Q-transformis the introduction of 2−1 in f Q(2−1x) which shifts the zeros of the polynomial f somehowand will have a large impact on the success of producing infinite sequences of irreduciblepolynomials.

From f R(x) = (2x)n f (2−1(x + 1/x)) it is seen that f R(x) is a self-reciprocal polynomialof degree twice that of f (x). Since f R(x) = 2n f Q(2−1x) we deduce from remark 4.3 thatf 7→ f R is an bijective mapping from the set of polynomials of degree n to the set of self-reciprocal polynomials of degree 2n. This mapping will occasionally be referred to asthe R-transform. Thus, every self reciprocal polynomial f (x) of degree 2n over Fq can bewritten f (x) = gR(x) for some g(x) ∈ Fq[x]. This will be used in the sequel. Furthermore,the R-transform of a product is the product of the R-transforms, so f R is irreducible onlyif f is.

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Since the R-transform of a monic polynomial f and the number of monic polynomialsof degree n equals the number of self-reciprocal polynomials of degree n we have that fis monic if and only if f R is monic.

The success of producing sequences of irreducible polynomials using the Q-transformrelied heavily on the connection between f (2) f (−2) and f Q(2) f Q(−2) and in general itis hard to say sensible things about this connection. To illustrate this, if we start withan irreducible polynomial f (x) over F11 where f (2) f (−2) is a quadratic non-residue,then in order to apply the Q-transform once more, we must assert that f Q(2) f Q(−2) =

(−4)deg f f (2 + 2−1) f (−2 + (−2)−1) = (−4)deg f f (3) f (−3) is a quadratic non-residue. Ingeneral, there need not be any connection between the non-quadratic nature of f (2) f (−2)and the properties of f (3) f (−3).

However, in the case of the R-transform, there is a rather clear connection betweenλ( f ) and λ( f R) on which our success of producing sequences of irreducible polynomialsis heavily dependent.

Note that with our new notation, theorem 4.1 states precisely that the sequence definedby fn+1(x) = f R

n (x) is a sequence of irreducible polynomials under certain conditions.The following lemmata constitute the proof of theorem 4.1 in Cohen’s paper [2]:

Lemma 4.20. If f is a polynomial over Fq, a finite field of odd characteristic p, deg f = n,then

• if p ≡ 1 mod 4 and if λ( f ) is a quadratic non-residue of Fq, then λ( f R) is aquadratic non-residue of Fq as well.

• if n is even, and if λ( f ) is a quadratic non-residue of Fq, then λ( f R) is a quadraticnon-residue of Fq as well.

Proof. λ( f R) = f R(1) f R(−1) = (−1)n22n f (1) f (−1) = (−1)n22nλ( f ). If n is even, thisclearly is a quadratic non-residue of Fq. If p ≡ 1 mod 4, −1 is a quadratic residue of Fq,so λ( f R) is a quadratic non-residue. �

Lemma 4.21. Let f be an irreducible polynomial over Fq, a finite field of odd characteris-tic, deg f = n. Then f R is irreducible over Fq if and only if λ( f ) is a quadratic non-residueof Fq.

Proof. Let g(x) = 2n f (2−1x), so that f R(x) = gQ(x) i.e. if we show that gQ(x) is irreduciblethen f R(x) is irreducible. g(x) is irreducible, for otherwise 2n f (2−1x) = r(x)s(x), forsome r(x), s(x) ∈ Fq[x] with 0 < deg r, s < n, and f (x) = 2−ns(2x)g(2x), contradictingirreducibility of f . By theorem 4.15, gQ(x) is irreducible if and only if g(2)g(−2) is aquadratic non-residue of Fq and g(2)g(−2) = 22nλ( f ) proves the claim. �

Those lemmata provide a proof of this version of theorem 4.1:

Theorem 4.22. Let f1(x) be a monic irreducible polynomial over Fq, a finite field of oddcharacteristic p, deg f1 = n, with n even if p ≡ 3 mod 4, and with λ( f ) a quadraticnon-residue of Fq. Then the sequence of polynomials defined by

fm+1(x) = f Rm (x), m ∈ N>0

is a sequence of monic irreducible polynomials over Fq, with deg fm = n2m−1.

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Remark 4.23. However, it is not the case that the theorem succeeds if the required even-ness of m if p ≡ 3 mod 4 is neglected, as will now be shown. Take x3 + 2 ∈ F7[x], whichis irreducible over F7, since it has no zeros in F7. Also λ( f ) = 3, which is a quadraticnon-residue of F7 since 33 = 27 = −1 in F7. Now f R(x) = x6 + 3x5 + 2x3 + 3x2 + 1,λ( f R) = 1, and furthermore

f R2(x) = x12 + 4x10 + 2x9 − x8 − x7 − x5 − x4 + 2x3 + 4x2 + 1

= (x6 + 3x5 + 3x3 + 2x2 − x − 2)(x6 + 4x5 − x4 + 2x3 + 2x + 3)

which is not irreducible!Actually, the theorem will invariably fail if the initial polynomial is of odd degree

when char Fq ≡ 3 mod 4. For if we start with f (x) ∈ Fq[x] which is irreducible over Fq

of odd degree m where λ( f ) is a quadratic non-residue of Fq then λ( f R) = (−1)n22nλ( f )which actually is a quadratic residue of Fq due remark 2.13.

The following theorem shows that every irreducible, self reciprocal polynomial ofdegree 2n arises by taking the R-transform of an irreducible polynomial of degree n onwhich theorem 4.22 can (possibly) be applied.

Theorem 4.24. Let f (x) ∈ Fq[x] be monic, self-reciprocal and irreducible of degree 2nover the finite field Fq. Then there is g(x) ∈ Fq[x], monic of degree n s.t. g(x) is irreducibleover Fq and

f (x) = gR(x), λ(g) a quadratic non-residue of Fq.

Proof. By remark 4.19 we find that f (x) = gR(x) for some monic g(x) of degree n sincethe R-transform is a bijective mapping from the set of polynomials of degree n to the setof self-reciprocal polynomials of degree 2n. In addition, the R-transform of a polyno-mial is irreducible only if the polynomial transformed is irreducible. Thus, g(x) must beirreducible. Furthermore, since

gR(x) = 2ng((2−1x))Q = (2ng(2−1x))Q := hQ(x)

where h(x) = 2ng(2−1x), is irreducible, we must have, by theorem 4.15, that h(2)h(−2) isa quadratic non-residue of Fq and since

h(2)h(−2) = 2ng(1)2ng(−1) = 22nλ(g)

we have deduced that λ(g) is a quadratic non-residue of Fq. �

From the previous theorem we can derive this corollary:

Corollary 4.25. Let Fq be a finite field, where charFq = p ≡ 1 mod 4. If f (x) ∈ Fq[x] isself-reciprocal and irreducible of even degree, then λ( f ) is a quadratic non-residue of Fq.

Proof. By theorem 4.24 we can find g(x) ∈ Fq[x] irreducible s.t. g(x)R = f (x) where λ(g)is a quadratic non-residue of Fq. By lemma 4.20 we find that λ(gR) = λ( f ) is a quadraticnon-residue of Fq. �

Now, at last, an example of when theorem 4.22 works:

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Example 4.26. Consider the polynomial x2 +2 ∈ F5, irreducible, and let α satisfy α2 +2 =

0 i.e. α2 = 3. Then F25 � F5(α). As can be verified, α has order 8 in F∗25 and 2 − α hasorder 3. Thus θ := α(2 − α) = 2(1 + α) has order 3 · 8 = 24 and generates F∗25.

We now attempt to find a polynomial f (x) of the form x + β where β ∈ F25 such thatλ( f ) = (1 +β)(−1 +β) is a quadratic non-residue of F25 so that theorem 4.22 is applicable.It turns out that β = θ + 1 is a good choice, since −1 + β = θ is a quadratic non-residueand 1 + β is a quadratic residue, as shown through the following calculation

(1 + β)(25−1)/2 = (2(α + 2))12 = . . . = 1.

Thus their product λ( f ) is a quadratic non-residue and theorem 4.22 applies: For instance

f R(x) = 2(2−1x + β)Q = 2x(2−1(x + 1/x) + β) = x2 + 2βx + 1 = x2 + (1 − α)x + 1

is irreducible over F25.We now show that theorem 3.4 can be applied to obtain a ”parallell” sequence of

irreducible polynomials to the sequence generated by f (x). As has been shown, f R(x) =

x2 + (1 − α)x + 1 is irreducible. Now, there is a non-trivial automorphism of F25 over F5,namely the one defined by σ(θ) = θ5. Hence

g(x) := σ( f R(x)) = x2 + σ(1 − α)x + 1 = x2 + (1 − α5)x + 1 = x2 + (1 + α)x + 1

is another irreducible (self reciprocal) polynomial of degree 2 over F25 by theorem 3.4.Now, by theorem 4.24 it holds that g(x) = hR(x) for some irreducible polynomial h(x)

satisfying λ(h) not a quadratic residue of F25. Since the R-transform is a bijective mappingfrom the set of polynomials of degree n to the set of self reciprocal polynomials of degree2n we must have h(x) , f (x) and thus h(x) can be used to generate a ”parallell” sequenceof polynomials to the sequence generated by f (x) through theorem 4.22.

It can indeed be calculated that

h(x) = x − 2 − 2α , f (x) = x − 2 + 2α

and thatλ(h) = (−1 − 2α)(2 − 2α) = 3α

is a quadratic non-residue, since (3α)12 = α12 = 36 = (32)3 = (−1)3 = −1.

Theorem 4.24 allows us to deduce that there is f (x) ∈ Fq[x] of degree n with λ( f ) aquadratic non-residue of Fq to which theorem 4.22 may be applied, provided that there is amonic self-reciprocal polynomial of degree 2n. The number of such polynomials is givenby formulae presented in [5], but here we will be content with presenting a polynomialthat exhibits properties regarding irreducible self reciprocal polynomials similar to thoseexhibited by xqn

− x regarding irreducible polynomials of degree d | n.

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4.4 The polynomial xqn+1 − 1

We will conclude the text by investigating the polynomial xqn+1 − 1 whose irreduciblefactors are in close kinship with irreducible self-reciprocal polynomials. We will see thatmost of the irreducible factors of xqn+1 − 1 are self-reciprocal and in order to show thiswe need the following two results on the identification and properties of self-reciprocalpolynomials:

1. Let f (x) ∈ Fq[x] be irreducible of even degree and let its set of zeros be closedunder inversion, i.e. if f (α) = 0 then f (α−1) = 0 for 0 , α in the splitting field off (x). Then f (x) is self-reciprocal.

Proof. From f ∗(x) = xdeg f f (1/x) we see that 0 = αdeg f f (1/α) = f ∗(α) so f (x) andf ∗(x) have the same set of zeros. Thus, since f is irreducible and hence has onlysimple zeros, we may write f (x) = c f ∗(x) for some c ∈ Fq. Using remark 2.15 wefind f ∗ = (c f ∗)∗ = c( f ∗)∗ = c f which gives c−1 = c so that c = ±1. Since the degreeof f is even of degree 2m say, there is a coefficient, namely that of xm, call it am,left unchanged when mapping f to f ∗. This means that we must have am = cam

so if am , 0 we must have c = 1. If am = 0 and c = −1 we get ak = −a2m−k for0 ≤ k ≤ 2m which implies f (1) = 0, contradiction to f (x) being irreducible. Thusc = 1 and f (x) = f ∗(x). �

2. If f (x) ∈ Fq[x] is irreducible and self-reciprocal with deg( f ) = m > 1, then m iseven.

Proof. Let α be a zero of f in the splitting field Fqm of f . Then, α , 0 since f isirreducible. Since 0 = f (α) = f ∗(α) = αn f (α−1), α−1 is also a zero of f . Sincewe can pair each zero α with another zero α−1 and we have α−1 , α (for otherwiseα = ±1) and since inverses are unique there must be an even number of distinctzeros of f in Fqm and so m is even. �

Now let Fq be a finite field and consider the polynomial

hq,n(x) = xqn+1 − 1 ∈ Fq[x].

We list some properties of hq,n(x):

• If α ∈ Fq is a zero of hq,n(x) then α ∈ {±1}:

αqn+1 − 1 = αqn−1α2 − 1 = α2 − 1 = 0

so α is a zero of x2 − 1.

• Let f (x) be irreducible and self reciprocal of degree 2n, then f (x) | hq,n(x). Let αbe a zero of f (x). Then the zeros of f (x) are {α, αq, . . . , αq2n−1

}and since f (x) is selfreciprocal there exists 1 ≤ j ≤ 2n − 1 s.t. α−1 = αq j

. We find that α is a zero of

h j(x) = xq j+1 − 1.

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For any polynomials xc − 1, xd − 1 we have xc − 1 | xd − 1 if c | d because

(y − 1)(yn−1 + yn−2 + . . . + y2 + y + 1) = yn − 1

for y in any commutative ring and n ∈ N. Apply with y = xc and n = d/c. Therefore

h j(x) | xq2 j−1 − 1 = x(q j+1)(q j−1) − 1

We now have that f (x) | h j(x) and h j(x) | xq2 j−1 − 1 which implies f (x) | xq2 j−1 − 1.Thus, as seen in the proof of theorem 2.2, we get 2n | 2 j and so n | j and thus n = jin other words hq,n(x) = h j(x) so we have shown f (x) | hq,n(x).

• Now let f (x) be an irreducible factor of hq,n(x) of degree m ≥ 2. Let α be a zero off (x). Since f (x) | hq,n(x) we have that αqn+1 = 1 and thus α−1 = αqn

. By theorem2.2 the element αqn

is a zero of f (x). Thus the set of zeros of f (x) is closed underinversion and by remark 2.15, f (x) is self reciprocal of even degree, m = 2d say.Since f (x) divides hq,n(x) it divides xq2n−1 − 1 as shown above and thus 2d | 2n andd | n.

What we can conclude from this information is that every irreducible factor of hq,n(x) =

xqn+1 − 1 of degree 1 is either x − 1 or x + 1 (they actually occur with multiplicity at most1, this can be shown by introducing the concept of the derivative of a polynomial, but willnot be done here, see [1] for instance). Any irreducible factor of degree higher than 1 ofhq,n(x) is of even degree and self reciprocal with degree dividing 2n.

The properties of hq,n(x) suggest that if one wants to find irreducible self reciprocalpolynomials of even degree one should investigate the divisors of hq,n(x).

What Meyn does in his paper [5] to determine a formula for the number of self-reciprocal irreducible polynomials of certain degree is very similar to what was done intheorem 2.21, namely to apply a Mobius inversion to a certain identity involving hq,n(x),in particular, there are self-reciprocal irreducible polynomials of degree 4 over any Fp

where p is prime congruent 3 modulo 4, which implies that there are suitable startingpolynomials for theorem 4.22 of degree 2.

The following argument shows that there are suitable starting polynomials of degree1 over Fp, p prime and p ≡ 1 mod 4:

We seek a polynomial f (x) = x + α s.t. λ( f ) is a quadratic non-residue of Fp i.e. s.t.λ( f ) = α2 − 1 is a quadratic non-residue. Thus, it suffices to find β ∈ Fp s.t. β = α2 isa quadratic residue but β − 1 = α2 − 1 is not. Suppose for a contradiction that there isno such β. Then, for every β that is a quadratic residue, β − 1 is a quadratic residue aswell. Hence the set {β − n : n ∈ N} must consist only of quadratic residues. However,{β − n : n ∈ N} = Fp which is a contradiction, since by remark 2.13 there are exactly(p − 1)/2 > 0 quadratic non-residues of Fp.

We are now guaranteed the existence of polynomials satisfying all requirements oftheorem 4.22 for any field Fp, p prime. Consequently, it is possible to generate infinitesequences of irreducible polynomials over Fp[x], p prime, which suffices if one wants tofind explicit descriptions of certain field extensions of Fp[x].

To conclude the text, we observe that even though we have no explicit starting poly-nomial for the application of theorem 4.22 in any given case we have a good candidatepolynomial hq,n(x) to look for such in its set of divisors. A task of this sort can be givento a computer and seems to be a rather small effort when the reward is an entire infinitesequence of irreducible polynomials!

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References[1] A. A. Albert Fundamental concepts of higher algebra, The University of Chicago

Press, 1956.

[2] S. D. Cohen The explicit construction of irreducible polynomials over finite fields,Designs, codes, and cryptography vol 2, 1992.

[3] R. Lidl, H. Niederreiter Finite Fields. Encyclopedia of Mathematics and its applica-tions 20, Cambridge University Press, 2008.

[4] Lars-Åke Lindahl Linjar Algebra, Fjarde upplagan, Matematiska institutionen, Up-psala Universitet, 2009.

[5] H. Meyn On the Construction of Irreducible Self-Reciprocal Polynomials Over Fi-nite Fields, Applicable algebra in engineering, communication and computing vol1, 1990.

[6] F. J. MacWilliams, N. J. A. Sloane The theory of error correcting codes, North-Holland, 1978.

[7] I. Niven, H. S. Zuckerman, H. L. Montgomery An introduction to the theory ofnumbers, Wiley, Fifth Edition, 1991.

[8] I. Shparlinski: Finite Fields. Theory and computation, Kluwer Academic Publishers,1999.

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